Question: The grades on a geometry midterm at Almond are normally distributed with $\mu = 79$ and $\sigma = 4.5$. Kevin earned a $78$ on the exam. Find the z-score for Kevin's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Kevin's exam grade by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{78 - {79}}{{4.5}}} $ ${ z \approx -0.22}$ The z-score is $-0.22$. In other words, Kevin's score was $0.22$ standard deviations below the mean.